The claim about () holds by the following arguments. Let \(F(x) = (f_1(x), \dots , f_s(x))\) . Under a sum rule (see, for example, [1]}), for \(x\in \operatorname{dom}f_0\) ,
\(\partial f(0,x) = \mathbb {R}^s \times \Big ( \mathop {\sum }\nolimits _{i=1}^s p_i \nabla f_i(x) + \partial f_0(x) \Big )\) 
